Helium gas at 300 K is shifted from a vessel of `250 cm^(3)` to a vessel of 1 L capacity. The pressure of gas will

Options

A. become double
B. becomes four times
C. decrease to half of the original value
D. decrease to one-fourth of the original value

Answer

To solve the problem of how the pressure of helium gas changes when it is transferred from a vessel of 250 cm³ to a vessel of 1 L (1000 cm³) at a constant temperature of 300 K, we can apply Boyle's Law. Here’s a step-by-step solution:

### Step 1: Understand Boyle's Law
Boyle's Law states that for a given mass of gas at constant temperature, the pressure (P) of the gas is inversely proportional to its volume (V). This can be mathematically expressed as:
\[ P_1 V_1 = P_2 V_2 \]
where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( P_2 \) = final pressure
- \( V_2 \) = final volume

### Step 2: Identify the Initial and Final Conditions
From the problem:
- Initial volume \( V_1 = 250 \, \text{cm}^3 \)
- Final volume \( V_2 = 1 \, \text{L} = 1000 \, \text{cm}^3 \)

### Step 3: Set Up the Equation
Using Boyle's Law, we can set up the equation:
\[ P_1 \times 250 = P_2 \times 1000 \]

### Step 4: Solve for Final Pressure
Rearranging the equation to solve for \( P_2 \):
\[ P_2 = \frac{P_1 \times 250}{1000} \]

### Step 5: Simplify the Equation
Simplifying the equation gives:
\[ P_2 = \frac{P_1}{4} \]
This indicates that the final pressure \( P_2 \) is one-fourth of the initial pressure \( P_1 \).

### Conclusion
Thus, when helium gas is transferred from a vessel of 250 cm³ to a vessel of 1 L, the pressure of the gas decreases to one-fourth of its original value.

### Final Answer
The pressure of the gas will be decreased to one-fourth of the original value.

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